# The Simplex Method in Tabular Form ```The Simplex Method in Tabular Form
In its original algebraic form, our problem is:
Maximize
Subject to:
z
z −4x1 −3x2
2x1 +3x2 +s1
−3x1 +2x2
+s2
2x2
+s3
2x1 +x2
+s4
=
=
=
=
=
0
6
3
5
4
(0)
(1)
(2)
(3)
(4)
x1 , x2 , s1 , s2 , s3 , s4 ≥ 0 .
Since the objective function and the nonnegativity constraints do not explicitly participate
in the mechanics of the solution procedure, we only need to present the coefficients in the
constraint equations. In tabular form, this problem will be represented as follows.
z
1
0
0
0
0
x1
−4
2
−3
0
2
x2
−3
3
2
2
1
s1
0
1
0
0
0
s2
0
0
1
0
0
s3
0
0
0
1
0
s4
0
0
0
0
1
0
6
3
5
4
Thus, each constraint equation is translated into a row of coefficients; and the coefficients
of a variable in different equations are listed in the same column, with the name of that
we have delimited the table into several areas, e.g., (i) the z-column at the left, (ii) the
coefficients in equation (0) at the top row, and (iii) the right-hand side constants of the
equations at the right-most column.
The above table will be referred to as the initial Simplex tableau. To simplify statements,
we will refer to the successive rows in the tableau as R0 , R1 , and so on; this numbering,
of course, corresponds to that of the original equations. In addition, we will refer to the
last column as the RHS column (since it comes from the right-hand-side constants in the
equations).
Associated with this initial tableau, the nonbasic variables are x1 and x2 and the basic
variables are s1 , s2 , s3 , and s4 . Therefore, the initial (or current) basic feasible solution is:
(x1 , x2 , s1 , s2 , s3 , s4 ) = (0, 0, 6, 3, 5, 4). This solution has an objective-function value 0,
which is the right-most number in R0 .
1
Consider R0 . Since the coefficients of x1 and x2 (the nonbasic variables) in that row are
both negative, the current solution is not optimal. Furthermore, since the coefficient of x1 ,
namely −4, is more negative than that of x2 , we will select x1 as the entering variable.
We will refer to the x1 -column as the pivot column. This terminology is suggested by the
fact that a round of Gaussian elimination is also called a pivot.
To determine the maximum possible increase in x1 , we conduct a ratio test. The ratio test
will involve the coefficients in the pivot column and in the RHS column. This is worked out
on the right margin of the tableau, as shown below.
Basic
Variable
s1
s2
s3
s4
z
1
0
0
0
0
x1
−4
2
−3
0
2
x2
−3
3
2
2
1
s1
0
1
0
0
0
s2
0
0
1
0
0
s3
0
0
0
1
0
s4
0
0
0
0
1
Ratio Test
0
6
3
5
4
6/2 = 3
−
−
4/2 = 2
← Minimum
Note that we did not compute a ratio for R2 and R3 , since both of these two rows have
a nonpositive coefficient in the pivot column (indicating that the corresponding equations
(2) and (3) do not impose any bound on x1 ). Since the minimum ratio appears in R4 , the
basic variable currently associated with that row, s4 (indicated at the left margin), will be
the leaving variable. We will refer to R4 as the pivot row.
With s4 leaving and x1 entering, the new basis will be x1 , s1 , s2 , and s3 . Therefore, we are
now interested in constructing a new tableau that is targeted to assume the configuration
specified below.
Basic
Variable
s1
s2
s3
x1
z
1
0
0
0
0
x1
0
0
0
0
1
x2
?
?
?
?
?
s1
0
1
0
0
0
s2
0
0
1
0
0
s3
0
0
0
1
0
s4
?
?
?
?
?
?
?
?
?
?
As before, the ?’s in the tableau represent blanks whose entries are to be determined.
To create this target tableau, we will employ row operations. As examples, the new row
4 will be generated by multiplying R4 by 1/2; and the new row 0 will be generated by
multiplying R4 by 2 and adding the outcome into R0 . Repeating similar operations for the
2
other rows yields the new tableau below.
2 &times; R4 + R0 :
(−1) &times; R4 + R1 :
(3/2) &times; R4 + R2 :
0 &times; R4 + R3 :
(1/2) &times; R4 :
z
1
0
0
0
0
x1
0
0
0
0
1
x2
−1
2
7/2
2
1/2
s1
0
1
0
0
0
s2
0
0
1
0
0
s3
0
0
0
1
0
s4
2
−1
3/2
0
1/2
8
2
9
5
2
Note that on the left margin of this tableau, we have explicitly indicated how individual
new rows are derived from those in the initial tableau. For example, the operations leading
to the new row 0 is listed as 2 &times; R4 + R0 , which corresponds to the earlier description.
In this round of Gaussian elimination, or pivot, the entry 2 located at the intersection of
the pivot column and the pivot row in the initial tableau plays a “pivotal role,” in that it
is repeated used to generate all five multipliers to R4 . We shall refer to this entry as the
pivot element.
The basis associated with the new tableau is: x1 , s1 , s2 , and s3 . Therefore, the new
basic feasible solution is: (x1 , x2 , s1 , s2 , s3 , s4 ) = (2, 0, 2, 9, 5, 0). This solution has an
objective-function value 8. Since there is a negative coefficient in the new R0 , namely the
−1 in the x2 -column, the current solution is not optimal. This completes the first iteration
of the Simplex method.
Next, since x2 is now the entering variable, the x2 -column is the new pivot column. To
determine the pivot row, we again conduct a ratio test, which is shown below.
Basic
Variable
s1
s2
s3
x1
z
1
0
0
0
0
x1
0
0
0
0
1
x2
−1
2
7/2
2
1/2
s1
0
1
0
0
0
s2
0
0
1
0
0
s3
0
0
0
1
0
s4
2
−1
3/2
0
1/2
Ratio Test
8
2
9
5
2
2/2 = 1
9/(7/2) = 18/7
5/2
2/(1/2) = 4
← Minimum
This shows that the new pivot row will be R1 , and the basic variable associated with that
row, s1 , will be the leaving variable.
With the entry 2 (Where is it?) as the pivot element, we now go through another pivot to
obtain the new tableau below.
(1/2) &times; R1 + R0 :
(1/2) &times; R1 :
(−7/4) &times; R1 + R2 :
(−1) &times; R1 + R3 :
(−1/4) &times; R1 + R4 :
z
1
0
0
0
0
x1
0
0
0
0
1
x2
0
1
0
0
0
s1
1/2
1/2
−7/4
−1
−1/4
3
s2
0
0
1
0
0
s3
0
0
0
1
0
s4
3/2
−1/2
13/4
1
3/4
9
1
11/2
3
3/2
Here again, the operations that led to this tableau are indicated on the left margin.
The basic feasible solution associated with this new tableau is (3/2, 1, 0, 11/2, 3, 0), with
a corresponding objective-function value of 9. Moreover, since the coefficients of s1 and s4
in the new R0 are positive, this solution is optimal. This completes the second iteration,
and the solution of this problem.
4
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